Question 422433
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Hi
hyperbola with vertices (0, –1) and (4, –1), and foci (–1, –1) and (5, –1).
Note: Line of symmetry is y = -1, Hyperbola opens right and left
Standard Form of an Equation of an Hyperbola is  {{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}} 
where Pt(h,k) is a center  with vertices 'a' units right and left of center.
 vertices (0,-1) and (4,-1) tell us the Center is (2,-1)
thus: {{{(x-2)^2/4 - (y+1)^2/b^2 = 1}}} 
foci are 3 units from center: "sqrt(4+5)" = 3,  b =5
 {{{(x-2)^2/4 - (y+1)^2/5 = 1}}} 
Always recommend sketching it out:
{{{drawing(300,300,   -6, 6, -6, 6,   grid(1),
circle(2, -1,0.3),
circle(-1, -1,0.3),
circle(5, -1,0.3),
graph( 300, 300, -6, 6, -6, 6,0,-1,2.236sqrt(((x-2)^2/4)-1)-1,-2.236sqrt(((x-2)^2/4)-1)-1))}}}